Measurement Pythagoras’ theorem - Spark Pythagoras’ theorem 6 ... 5aIf p2 = 36, ﬁnd p if p is...

Measurement

Pythagoras’ theorem

6

Pythagoras lived around 500 BC. The theorem (or rule) which carries his name was well known before this time, but had not been proved (as far as we know) until Pythagoras proved it. Over 300 proofs for the theorem are known today. The theorem, which describes the relationship between the side lengths of a right-angled triangle, is used in fields as diverse as construction, electronics and astronomy.

■ calculate squares and square roots of numbers■ identify the hypotenuse as the longest side of any right-angled triangle■ establish the relationship between the length of the sides of a right-angled

triangle in a practical way■ use Pythagoras’ theorem to find the length of the hypotenuse or one of

the shorter sides in a right-angled triangle■ use Pythagoras’ theorem to establish whether a triangle has a right angle■ use Pythagoras’ theorem to solve practical problems involving right-

angled triangles■ graph two points to form an interval on the number plane and form a

right-angled triangle with that interval as the hypotenuse■ use Pythagoras’ theorem to calculate the length of an interval on the

number plane.

■ hypotenuse The longest side of a right-angled triangle. It is the side opposite the right angle.

■ theorem A mathematical rule or law that can be proved.■ diagonal An interval from one vertex (corner) of a shape to another non-

adjacent vertex.■ interval A section of a line, having a definite length.■ exact length A length that is written as an exact value, and is not rounded

to decimal places.■ Cartesian plane Another name for the number plane.

What is the maximum length of a steel rod that will fit inside a rectangular box whose base dimensions are 70 cm × 50 cm and whose height is 190 cm? How could the maximum length of the rod be determined?

In this chapter you will:

Wordbank

Think!

PYTHAGORAS’ THEOREM 217 CHAPTER 6

218 NEW CENTURY MATHS 9

Squares and square rootsThe square of a number is that number multiplied by itself. For example, the square of 5, written 52, is 5 × 5 = 25.

The square root of a number is the number which, when multiplied by itself, results in the given

number. For example, the square root of 9 (written as ) is 3, because 3 × 3 = 9.

1 Convert each of these lengths to metres:a 500 cm b 2000 mm c 4 kmd 6000 mm e 150 cm f 1200 cmg 6.7 km h 1800 mm i 85 cm

2 Evaluate:a 72 b 122 c 92 + 42

d e 82 + 62 f

g 122 − 42 h i

j k l

3 Calculate, correct to one decimal place:

a b c

d e f

4 Solve these equations by finding the value of the pronumeral each time:a 40 = c + 27 b 18 = y + 6 c 24 = m − 16 d 145 = y − 20

5 a If p2 = 36, find p if p is positive. b If k is positive, and k2 = 100, find k.

6 Write the coordinates of the points A, B, C … L on this number plane:

121 576

100 576+ 372

352

–

2( )2 7( )2 502

402

–

75 160 94

200 787 22 500

O

A

B

C

D

E

F

G

H

I

J

K

L

−4 −3 −2 −1 1 2 3 4

4

3

2

1

−1

−2

−3

−4

5

y

x

Start up

Worksheet 6-01

Brainstarters 6

9

PYTHAGORAS’ THEOREM

219

CHAPTER 1CHAPTER 6

Example 1

Find the value of:

a

17

2

b

9.7

2

c

4

2

+

7

2

Solution

Find the value (correct to three significant figures, where necessary) of:

a b c

Solution

a 172 = 289

17 (on a calculator)

b 9.72 = 94.09

9.7 (on a calculator)

c 42 + 72 = 16 + 49 = 65

4 7 (on a calculator)

a = 32


b = 9.110 433 ≈ 9.11


c = 8.602 325 … ≈ 8.60

5 7

or 5 7 (on a calculator)

x2 = x2 =

x2 + x2 =

Example 2

1024 83 52 72+

1024

√ =83

√ =

52 72+

x2 + x2 = √ =

√ ( x2 + x2 ) =

1 Find the value of:a 72 b 32 c 152 d (6.2)2

e (31.7)2 f ( )2 g ( )2 h ( )2

i (0.1)2 j (0.71)2 k 32 + 52 l 202 − 72

m 92 + 402 n 52 + 182 o 502 − 402 p (3.2)2 − (1.5)2

q (1.2)2 + (0.7)2 r (0.3)2 + (0.4)2 s (2 )2 + 32 t ( )2 + ( )2

2 Find the value of each of the following (correct to two decimal places where necessary):

a b c d

e f g h

i j k l

m n o p

q r s t

12--- 2

5--- 5

4---

12--- 1

4--- 1

5---

144 25 8116------ 1.21

15 900 170 42.7

0.04 24 7.8 32 42+

202 152– 12---( )

2 34---( )

2+ 0.6( )2 0.1( )2+ 4.5( )2 32–

52---( )

2 32---( )

2– 8.2( )2 3.5( )2– 4.1( )2 0.9( )2– 21

2---( )2 12+

Exercise 6-01Example 1

Example 2


TrianglesNaming the sides in a triangleFor triangle ABC the lengths of the sides of the triangle are usually named in the following manner:• the side opposite ∠A is called a• the side opposite ∠B is called b• the side opposite ∠C is called c

The hypotenuseIn any right-angled triangle, the longest side is called the hypotenuse.

The hypotenuse in the diagram is AB (or BA) , or c because it is opposite the right angle ∠C.

Rope stretchers

In ancient Egypt most of the farms were built along the Nile River. Whenever the Nile flooded, the farmers were faced with rebuilding their fences. They could make the straight sections of fencing quite simply, but needed a method to construct the right angles at the corners of the fields. They used a technique established by a group of Egyptians called rope stretchers.

The rope stretchers took a length of rope with thirteen knots tied in it at equal intervals, as shown.

The rope was fixed to the ground at the fourth and eighth knots, and stretched around so that the first knot aligned with the thirteenth knot, as shown. The right angle is formed at the fourth knot.

The Egyptian farmers could not understand why this procedure produced a right angle. The reason was eventually discovered by the Greek mathematician Pythagoras (580–496 BC).

1 2 3 4 5 6 7 8 9 10 11 12 13

Just for the record

13th knot th knot

th knot

knot

units

unitsunits

2

33

4

4

5

5

6

7

89101112

st1

Worksheet 6-02

Pythagoras’ discovery

Worksheet 6-03

A page of right-angled triangles

A

B

C

a

b

c

AB

C

hypotenuse c

PYTHAGORAS’ THEOREM 221 CHAPTER 1CHAPTER 6

1 Name the hypotenuse in each of the following right-angled triangles.

2 Measure the hypotenuse in each of these triangles. Give your answer in millimetres (mm).

MN

P

V

T

Z

a

b

c

v

w

x

H G

N C

QT

r

p

q e

g

k

P T

R

V

M

P

d g

kn

w

p

a b c d

e f g h

i j k

l

a b

c

d

ef

g

h

i

j

Exercise 6-02


Working mathematicallyCommunicating: Hidden hypotenusesWork in pairs to complete this investigation.

The square ABCD, shown below, has been divided into many shapes, some of which are right-angled triangles.a List as many hypotenuses as you can find in 10 minutes, writing your answers in the

form ‘AC in ∆ABC’.b Share your list with the whole class to find all the hidden hypotenuses.

A B

CD

F

G

HJ

K

LM N

P

Q

R

S

I

Working mathematicallyApplying strategies and reasoning: A rule relating the sides of a right-angled triangleYou will need: a protractor, a calculator, a ruler and a pencil.

Step 1: In the centre of a new page accurately construct the right-angled triangle ABC shown below.(Diagrams not to scale.)

A

B C

3 cm5

4 cm

cm


What is Pythagoras’ theorem?What Pythagoras discovered about the sides of a right-angled triangle is a rule called Pythagoras’ theorem. A theorem is a mathematical rule or law that can be proved.

Pythagoras’ theorem is as shown in the box below.

Step 2: Construct a square on each side of the triangle, as shown.

1 a Calculate the area of each square you constructed.

b Add together the areas of the squares on the two shorter sides.

c Compare your result from part b to the area of the square on the hypotenuse.

d Describe the pattern 32 + 42 = …

2 Use your calculator to verify that the result you described in Question 1 part d is true for these right-angled triangles:

3 cm5

4 cm

cmArea X

Area Z

Area Y

61

60

40

85

75

2745

36

84

13

85

mmmm

mm

mmmm

mm

cm

cmcm

m

m

m

11

a b

c d

Worksheet6-02

Pythagoras’ discovery

In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

As a formula, this is written:

c2 = a2 + b2 A

B

C

a

b

c

Worksheet6-03

A page of right-angled

triangles


Example 3

State Pythagoras’ theorem for triangle XYZ, shown on the right.

SolutionPythagoras theorem for ∆XYZ is z2 = x2 + y2

Choose the correct statement of Pythagoras’ theorem for triangle PQR shown at the right:A PR2 = PQ2 + QR2

B QR2 = RP 2 + PQ2

C PQ2 = RP 2 + QR2

SolutionThe correct statement is statement B, (that is, QR2 = RP 2 + PQ2).

Choose the correct statement of Pythagoras’ theorem for ∆ABC shown.A 612 = 112 + 602

B 602 = 612 + 112

C 112 = 602 + 612

D 60 = 11 + 61

SolutionThe correct statement is statement A (that is, 612 = 112 + 602).

X

Y

Z

x

y

z

Example 4

P

Q

R

6061

11

Example 5

1 State Pythagoras’ theorem as it applies to each of the following right-angled triangles.

M K

N

k m

n

C

D

Pc

dp

K H

T

h k

t

D E

F

d

f

q

A B

C

5

1213

a b

W

X

Y

w

x

y

c

d e f


SkillBuilder 21-02



2 Choose the correct statement of Pythagoras’ theorem for each triangle shown below.a A y2 = x2 + w2 b A e2 = f 2 + g2

B x2 = w2 + y2 B g = e + fC w2 = x2 + y2 C f 2 = e2 + g2

D y = x + w D g2 = e2 + f 2

c A PR = PQ + QR d A a2 + c2 = b2

B QR2 = PQ2 + PR2 B c2 = a2 + b2

C RP2 = RQ2 + PQ2 C a2 = b2 + c2

D PQ2 = PR2 + QR2 D b2 = a2 + c2

e A KL2 + KJ2 = LJ2 f A f 2 = e2 + h2

B JL2 + KL2 = KJ2 B h2 = f 2 + e2

C KL2 = JK2 + JL2 C e2 = f 2 + h2

D K + J + L = 0 D FH2 = EF2 + EH2

g A q2 = n2 + t2 h A VK2 + VJ2 = JK2 B TN2 + TQ2 = NQ2 B j2 = v2 + k2

C n2 = q2 + t2 C v = j2 + k2

D TN2 = TQ2 + NQ2 D VK2 = JV2 + JK2

i A a2 = b2 + c2 j A x2 = m2 + p2

B b2 = a2 + c2 B w2 = q2 + x2

C c2 = a2 + b2 C m2 = q2 + w2

D none of these D x2 = q2 + w2

3 Choose the correct statement of Pythagoras’ theorem for each triangle shown below:

a A 742 + 702 = 202 b A 30 = 18 + 24B 242 + 702 = 742 B 182 + 302 = 242

C 70 = 20 + 74 C 182 = 242 + 302

D 202 + 742 = 702 D 302 = 182 + 242

c A 102 + 82 = 62 d A 152 + 132 = 202

B 10 + 8 = 6 B 52 + 122 = 132

C 62 + 82 = 102 C 52 + 122 = 202

D 6 + 8 = 10 D 122 + 152 = 202

w

x

y

e

f

g

P Q

R

a

b

c

KL

J

EF

H

QT

N

J

V K

a b

c

m

p

q

w

x

70

74

24

24

18

30

10 8

6

15

13

125

20

Example 4

Example 5


Working mathematicallyApplying strategies and reasoning: Proving Pythagoras’ theorem

You will need: tracing (or blank) paper, scissors, glue.

How can you prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides?

Proof 1One way is to cut out the squares on the two shorter sides and assemble them to completely cover the square on the hypotenuse.

Step 1: Make two copies of Figure A.

Step 2: Cut out the first copy of Figure A and paste it into your workbook, keeping all the squares intact.

Step 3: Cut out the second copy of Figure A. Colour the squares if you wish. Then cut the squares on the two shorter sides into the five numbered pieces shown.

Step 4: Arrange the five pieces to exactly cover the square on the hypotenuse. Paste this proof of Pythagoras’ theorem into your workbook as well.

Proof 2Step 1: Make two copies of Figure B,

on grid paper.

Step 2: Cut out the first copy of Figure B and paste it into your workbook, keeping all the squares intact.

Step 3: Cut out the second copy of Figure B. Then cut up the squares on the two shorter sides as shown. The pieces 2, 3, 4 and 5 are all the same size and shape as the triangle in the centre.

Step 4: Arrange the pieces to exactly cover the square on the hypotenuse. Paste this proof of Pythagoras’ theorem into your book as well.

2

1

4

5

3

Figure A

1

2 3

4

5

Figure B

Worksheet 6-04

Proving Pythagoras’

theorem

Geometry 6-01

Measuring Pythagoras

Geometry 6-02

Pythagoras dissection

Geometry 6-03

Perigal’s proof

Geometry 6-04

Leonardo’s proof


Finding the length of the hypotenusePythagoras’ theorem can be used to find the length of the hypotenuse of a right-angled triangle, given the lengths of the other two sides.

Pythagoras and his secret societyPythagoras’ theorem is the most famous mathematical rule because it is simple, used widely and, according to the Guinness Book of Records, it is the rule that has been proved in the greatest number of different ways. A book published in 1940 contained 370 proofs, including one by US president James Garfield (1831–1881), and more have been discovered since.

Pythagoras was a Greek mathematician and philosopher who lived from 580 BC to 496 BC. He did not discover the theorem that we know by his name, as it was already known to ancient Babylonian, Hindu and Chinese mathematicians. However, Pythagoras may have been one of the first to prove the theorem.

In 540 BC, Pythagoras formed a school called the ‘Pythagorean brotherhood’, a secret society involved with philosophy, religion and mathematics. The Pythagoreans were vegetarians and believed in reincarnation, following a strict code of conduct and living in communes without possessions. The group included women, and all members were treated equally. The symbol of the brotherhood was a pentagram, a five-pointed star formed by the diagonals of a regular pentagon. Many rules attributed to Pythagoras were actually discovered by his followers.

The Pythagoreans believed that numbers had magical properties, that all objects were made up of numbers and that ‘God is a mathematician’. The school motto was ‘All is Number’. Pythagoras and his followers found mathematical patterns in music and in astronomy.

Draw a pentagram, the symbol of the ‘Pythagorean brotherhood’, using the diagonals of a regular pentagon.

Just for the record

Skillsheet 6-01

Pythagoras’ theoremExample 6

Find the length of the hypotenuse in ∆ABC shown on the right.

Solutionc2 = a2 + b2

= 82 + 152

= 64 + 225

c2 = 289

c =

= 17

∴ The length of the hypotenuse is 17 m.

In ∆PQR, ∠P = 90°, PQ = 25 cm and PR = 32 cm. Sketch the triangle and find the length of the hypotenuse, p, correct to one decimal place.

A

B

C

8 m

m

m

c

15

289

Example 7

Spreadsheet 6-01



Solutionp2 = q2 + r2

p2 = 322 + 252

p2 = 1649

p = p = 40.607 881 01 …

≈ 40.6∴ The length of the hypotenuse is 40.6 cm (correct to one decimal place).

Find x in the diagram on the right.

SolutionSince the length of x is required in metres, change 24 cm to 0.24 m.x2 = 0.242 + 1.432

= 0.0576 + 2.0449x2 = 2.1025x =

= 1.45∴ The length of the hypotenuse is 1.45 m.

P

Q

Rq cm = 32 cm

r cm = 25 cmp cm

1649

Example 8

x m

24 cm

1.43 m

2.1025

1 Find the length of the hypotenuse in each of these triangles.

CB

A

X Y

Z

R

TV

A

B

C

T

WP

H

L

K

PQ

R

KP

J

A

B

C

8

6

12

35

12 16

5

12

24

10

4 m

4.2 m

18

24

30

16

7

24

cm

cm

cm

cm

cm cm

cm

cm

cm

cm

cm

cm

m

m

mm

mm

B

AC

H I

JD

F

L

9

40 m

35

1218

7.5

mm

mm

m

m

m

a b c

e f

g h i

j k

d

l



2 Find the length of the hypotenuse of each of these triangles, correct to one decimal place.

3 Find the value of the pronumeral in each of the following triangles (correct to one significant figure).

X

ZY

R

P

Q

K

L

N

C

D

E

V Z

T

AB

CQ

RT

R N

Q

Z

VG

10 cm

15 cm

11 cm

16 cm

6 cm

9 cm

19 cm

24 cm

54 mm25 mm

57 mm

34 mm

51 mm

39 mm

41 mm

72 mm

67 mm

84 mm

P

VF

T L

H

R T

V

49 mm

70 mm

1.76 m

1.85 m

2.4 m

6 m

a b c

d e

j k l

f

g h i

a b c

d ef

g h i

k mm

p mm

c mm

y cm

d m

x m a cm

q m

t m

45 mm

24 mm

45 cm

28 cm 40 m

30 m

75 mm

40 mm

0.5 m

1.2 m 80 cm

18 cm

2.1 m

7.2 m 9.1 m 6 m30 mm

16 mm

Example 8


4 Find the value of each pronumeral, correct to one decimal place.

5 For each of the following, sketch the triangle and then calculate the length of the hypotenuse. Make sure your answer is expressed in the correct units.a In ∆ABC, ∠ABC = 90°, AB = 39 cm, and BC = 57 cm. Find AC in centimetres, correct to

one decimal place.b In ∆MPQ, ∠PQM = 90°, QM = 2.4 m, and PQ = 3.7 m. Find PM in metres, correct to one

decimal place.c In ∆RVJ, ∠J = 90°, JV = 127 mm, JR = 42 mm. Find RV in millimetres, correct to the

nearest millimetre.d In ∆EGB, EG = EB = 127 mm, and ∠GEB = 90°. Find the length of BG in millimetres,

correct to one decimal place.e In ∆NKL, ∠K = 90°, NK = 75 cm, and KL = 1.2 m. Find NL in metres, correct to one

decimal place.f In ∆VZX, ∠V = 90°, VX = 247 mm, and VZ = 30.6 cm. Find ZX in millimetres, correct to

the nearest millimetre.g In ∆PQR, ∠RPQ = 90°, PQ = 2.35 m, and PR = 2.2 m. Find QR in metres, correct to two

decimal places.h In ∆DEF, ∠F = 90°, DF = 84 cm, and EF = 1.45 m. Find DE in centimetres, correct to the

nearest whole number.i In ∆PQR, ∠QRP = 90°, PR = 11.6 cm, and RQ = 9.7 cm. Find PQ in millimetres, correct

to the nearest whole number.j In ∆YTM, ∠T = 90°, TM = 3.4 m, and TY = 2.75 m. Find YM correct to three decimal

places.

a b c

d e f

g h i

k cm

g m

h m

y m

p m

a cm

f m

x cm

d m

10 cm

9 cm8 m

14 m

18 m

29 m

8 m

20 m

20 m

22 m

7 cm

12 cm

7.2 m

4.8 m 9.5 cm

8.4 cm

53 m

33 m

Example 7


Finding the length of a shorter sidePythagoras’ theorem can also be used to find the length of a shorter side of a right-angled triangle.

Spreadsheet activity: Finding the hypotenuse

1 Set up your spreadsheet as follows:

• The labels a, b, and c represent the length of the sides of a right-angled triangle, with c being the hypotenuse.

• To obtain values for c in rows 3, 4, …, fill down from C2.

2 Use your spreadsheet to find the length of the hypotenuse given the following pairs of values for the other two sides:a a = 5, b = 12 b a = 8, b = 8 c a = 12, b = 7d a = 46, b = 22 e a = 30, b = 25 f a = 18, b = 12g a = 80, b = 18 h a = 8, b = 4 i a = 10, b = 1

3 Print your results and paste them into your workbook.

A B C

1

2

3

4

�

a b c

= sqrt(A2^2 + B2^2)

Using technology

Spreadsheet

Example 9

Find the value of the pronumeral in the diagram on the right.

Solution172 = d2 + 82

∴ d2 + 82 = 172 (placing d2 on the left-hand side)d2 = 172 − 82 (subtracting 82 from both sides)

= 289 − 64d2 = 225d =

= 15∴ The value of d is 15.

Find x (correct to one decimal place).

d m

17 m8 m

225

Example 10

x m

8 m

3 m

Skillsheet 6-01



Solution82 = x2 + 32

x2 + 32 = 82

x2 = 82 − 32

= 64 − 9x2 = 55x =

= 7.41619 …x ≈ 7.4 ∴ The length of x is 7.4 m.

In ∆ABC, ∠C = 90°, AC = 36 mm, and AB = 60 mm. Find the length of BC.

SolutionSketch ∆ABC from the given information.

c2 = a2 + b2 (Pythagoras’ theorem)602 = a2 + 362 (substituting given information)

a2 + 362 = 602 (placing a2 on the left-hand side)∴ a2 = 602 − 362 (subtracting 362 from both sides)

a2 = 3600 − 1296a2 = 2304a =

= 48∴ The length of BC is 48 mm.

55

Example 11

A

BCa mm

36 mm60 mm

2304

1 Find the value of each pronumeral:

a b c

d e f

g h i

j k l

a m

x m

a mm

b m

p m

x cm

a mmx cm x cm

w mmx cm

p m

41 m9 m

63 m

65 m61 mm 11 mm

35 m

37 m

0.8 m1.7 m

6.3 cm 10.5 cm

12 mm

15 mm

8.4 cm

8.5 cm

72 cm

75 cm

25 mm

65 mm

170 cm

150 cm

63 m

65 m


SkillBuilder 21-16


Worksheet 6-05

Finding a missing side


2 Find the length of the unknown side in each of these triangles. Give your answers correct to one decimal place.

3 Find the length of the unknown side of each of these triangles. Give your answers correct to one decimal place.

a b c

d e f

g h i

j k l

x cm x cm x cm

a m

a m

w m

y cm

a m

e cm

x m

p cm

g m

27 cm

20 cm

16 cm

7 cm

75 cm

120 cm

32 m 21 m25 cm

43 cm

58 m

32 m

45 cm

84 cm

127 m103 m

50 m

62 m

3.7 cm4.9 cm

4.2 m

1.9 m

204 m 67 m

a b c

d e f

g h i

j k l

m cm k m

g mm

p cm

x mm

y m

t cm a m

c m

r m

q cmb m

17 cm

27 cm4 m

2.4 m

96 mm72 mm

21.6 cm

18.0 cm

32 mm17 mm

19.5 m34.1 m

5.1 cm

6.8 cm

2.1 m

0.8 m

44 m18 m

1.9 m

3.5 m

18 cm

13.7 cm

9.8 m

9.8 m

Example 10

Example 11


4 For each of the following, sketch the triangle and then calculate the length of the unknown side. Make sure your answer is expressed in the correct units.a In ∆ABC, ∠C = 90°, CA = 57 mm, and AB = 80 mm. Find BC correct to the nearest

millimetre.b In ∆PQR, ∠RPQ = 90°, PQ = 20 cm, and QR = 55 cm. Calculate RP correct to one

decimal place.c In ∆FHJ, ∠JHF = 90°, HJ = 1.3 m, and FJ = 1.65 m. Find FH in metres, correct to one

decimal place.d In ∆XYZ, ∠X = 90°, YZ = 2.4 m, and XZ = 1.3 m. Calculate XY in centimetres, correct to

the nearest centimetre.e In ∆PTV, ∠P = 90°, TV = 230 cm, and PV = 67 cm. Find PT in centimetres, correct to one

decimal place.f In ∆GKL, ∠LGK = 90°, KL = 527 mm, and GK = 21.6 cm. Find GL in millimetres, correct

to the nearest whole number.g In ∆ZXY, ∠X = 90°, ZY = 1.07 m, and XY = 234 mm. Find XZ in millimetres, correct to one

decimal place.h In ∆SPV, ∠V = 90°, PS = 190 cm, and PV = 84 mm. Find SV in centimetres, correct to one

decimal place.

Pythagoras’ theoremYou can solve problems involving Pythagoras’ theorem by using a graphics calculator. The formula for Pythagoras’ theorem is written as H2 = A2 + B2, where H is the hypotenuse, and A and B are the other two sides. The formula needs to be arranged to the form 0 = A2 + B2 − H2.

Casio CFS-9850GB PLUS

Step 1: Move to EQUA on the main menu and press EXE. Select F3 Solver.

Step 2: Next to ‘Eq:’ type A2 + B2 − H2.

Step 3: Type in the values of the variables you know and type 0 for the value you don’t know.

Step 4: Press F5 to work out the unknown.

Texas Instruments TI-83

Step 1: Press MATH 0 (Solver).

Step 2: Next to ‘eqn: 0 =’ type A2 + B2 − H2, and press ENTER. You may need to use the ‘up’ arrow to show the equation. The screen will show the variables.

Step 3: Type in the values you know and type 0 for the value you don’t know.

Step 4: Press ALPHA ENTER (SOLVE) to work out the unknown.

1 Use your calculator to answer the questions in Exercise 7-05.

Using technology


Spreadsheet activity: Finding the length of a shorter sideFor a right-angled triangle with sides, a, b, and c (where c is the hypotenuse), a shorter side can be found by using the formula:

b =

1 Set up your spreadsheet as shown below:

2 Use your spreadsheet to find the length of one of the shorter sides in a right-angled triangle given the following pairs of values for the other two sides:a a = 7, c = 15 b a = 6.8, c = 10.9 c a = 84, c = 100d a = 2, c = 4 e a = 22, c = 40 f a = 0.8, c = 1.8g a = 5.7, c = 15.8 h a = 9.5, c = 15.5

3 Print your results and paste them into your workbook.

c2 a2–

A B C

1

2

3

�

c a b

= sqrt(A2^2 − B2^2)

Using technology

Spreadsheet

Working mathematicallyReasoning: Pythagorean triadsA Pythagorean triad (or triple) is a group of three numbers that obey Pythagoras’ rule (the square of the largest number is equal to the sum of the squares of the other two numbers).

Consider the numbers, 5, 12 and 13. The number 13 is the largest number.

132 = 169

52 + 122 = 25 + 144

= 169

So, 52 + 122 = 132, and this means that (5, 12, 13) is a Pythagorean triad.

1 Show that each of the following sets of numbers is a Pythagorean triad.a (11, 60, 61) b (8, 15, 17)c (72, 96, 120) d (7, 24, 25)

2 Which of the following sets of numbers are Pythagorean triads? Check your answers with those of other students.a (36, 48, 60) b (13, 60, 61) c (21, 72, 75)d (9, 40, 43) e (9, 12, 15) f (9, 40, 41)

3 a The group (3, 4, 5) is a Pythagorean triad. Are multiples of (3, 4, 5) also Pythagorean triads? Explain your answer.

b Use the Pythagorean triad (5, 12, 13) to form three other triads. Compare your results with those of other students.


Testing for right-angled trianglesIf a triangle with sides of length a, b, and c, as shown, is right-angled, then we know, by Pythagoras’ theorem, that c2 = a2 + b2.

Conversely, for a triangle with sides of length x, y and z, as shown, if x2 + y2 = z2 then we know that the triangle must be right-angled. (The right angle will be opposite the hypotenuse, z.)

Expressing this rule another way, if the side lengths of a triangle form a Pythagorean triad, then the triangle must be right-angled.

Worksheet 6-06

Pythagorean practice

a

bc

x

y

z

x

y

z

Example 12

1 Is the triangle shown on the right a right-angled triangle?

Solution212 + 722 = 441 + 5184

= 5625752 = 5625

Since 212 + 722 = 752, the triangle is right-angled. (The right angle is ∠B).

2 Dan was asked to construct a right-angled triangle. The dimensions of his triangle were 37 cm, 12 cm and 40 cm. How do you know these measurements do not form a right-angled triangle?

SolutionIf Dan’s triangle is right-angled, then 122 + 372 = 402.But: 122 + 372 = 144 + 1369

= 1513and: 402 = 1600Since 122 + 372 ≠ 402, the triangle is not right angled.

A

BC

75 mm 21 mm

72 mm

1 From the measurements shown, test which of the following triangles are right angled. For each right-angled triangle, copy the triangle and mark the position of the right angle.

119 mm

105 mm

56 mm

12 cm

20 cm

16 cm

9 cm

50 cm67 cm

a b c



Pythagoras’ theorem in quadrilaterals

21 mm

72 mm75 mm

125 mm

120 mm

93 mm

60 cm

25 cm65 cm

d e f

g h i

j k l

14 cm9 cm

10 cm

2 m

2.7 m 3.45 m7.4 cm

5 cm

9.5 cm

1.13 m1.12 m

15 cm

264 cm

23 cm 265 cm

13.923 m

236 mm1392.5 cm

Example 13

1 How long is the diagonal AC in this rectangle?

SolutionLet x be the length of the diagonal.x2 = 682 + 512

= 4624 + 2601 x2 = 7225x =

= 85∴ The diagonal, AC, is 85 mm long.

2 Find the width, CD of this rectangle (correct to one decimal place).

SolutionLet CD be x m.

122 = x2 + 82

x2 + 82 = 122

x2 = 122 − 82

= 144 − 64x2 = 80x =

= 8.944 27 …≈ 8.9

∴ The width, CD, of the rectangle is 8.9 m.

68 mm

51 mm

A B

CD

x mm

7225

A

B C

D

12 m

8 m

x m

80

Worksheet6-06

Pythagorean practice


Example 14

Find the length of RQ in this trapezium, correct to one decimal place.

SolutionForm the right-angled triangle RZQ, as shown below, and let RQ be x cm.By comparing sides:

ZQ = 72 − 60 = 12 cmRZ = MP = 47 cm

∴ x2 = 472 + 122

= 2209 + 144x2 = 2353x =

= 48.507 731 34 …≈ 48.5 ∴ The length of RQ is 48.5 cm (correct to one decimal place).

Find the length of RT in this quadrilateral:

SolutionFind the length of VT first and then use it to find the length of RT .Let VT be q mm.q2 = 202 + 482 (using Pythagoras’ theorem in ∆VTQ)

= 400 + 2304q2 = 2704q =

= 52∴ The length of VT is 52 mm.

Now use VT = 52 mm and Pythagoras’ theorem in ∆VTR. Let RT be y mm.652 = 522 + y2

4225 = 2704 + y2

4225 − 2704 = y2

1521 = y2

y = = 39

∴ The length of RT is 39 mm.

MP

RQ

72 cm

47 cm

60 cm

MP

QR

Z

60 cm60 cm

47 cm

12 cmx cm

2353

Example 15

Q

R

T

V65 mm

48 mm

20 mm

Q

T

V

48 mm

20 mm

q mm

2704

R

T

V

52 mm

65 mm

y mm

1521

1 Find the size of the pronumeral in each of these quadrilaterals. Give your answer correct to two decimal places where necessary.

a b

35 mm

84 mm

7.5 cmx mm x cm

c

6.4 cm

12 cm

x cm



2 a A rectangle, PQRT, has side lengths of 75 mm and 100 mm. Find the length of the diagonal PR.

b A square has a side length of 96 mm. Find the length of its diagonals, correct to the nearest millimetre.

c A rectangle, ZVXY, has side lengths of 184 mm and 230 mm. Find the length of the diagonal ZX (correct to the nearest millimetre).

d The diagonal of a rectangle is 200 mm in length. One of the side lengths of the rectangle is 167 mm. Find the length of the other side, correct to the nearest millimetre.

e A 40 cm piece of wire is bent to form a square. How long is the diagonal of the square (correct to the nearest centimetre)?

3 Find the length of each unknown side of these quadrilaterals. Give your answer correct to one decimal place where necessary.

d e f38 cm

15 cm 9 cm

5 cm

18 cm

13 cmw cm

y cm

x cm

g h

111 mm

p mm105 mm 33 cm

h cm

i

120 cm

x cm

a b

d e f

A

B

C

D

AB

C

D

T

S

P

RZ

V

FP

L

M

N

P

39 cm

12 cm

43 cm

7.8 m

14 m

19 m

cM P

Q

R

2.5 m

2 m1.1 m

104 mm200 mm

278 mm

298 mm

202 mm100 mm

140 cm

61 cm

80 cm

Example 14


Practical applications of Pythagoras’ theoremWhen using Pythagoras’ theorem to solve problems, it is often useful to follow the steps below.• Read the problem carefully.• Draw a diagram and label any given information. Choose a variable to represent the length you

want to find.• If needed, make a separate sketch of the right-angled triangle you will use for calculations.• Use Pythagoras’ theorem to calculate the length.• Answer the question.

4 Find the value (in centimetres) of each pronumeral in the following quadrilaterals. Give your answer correct to one decimal place where necessary.

a A

B

C

Dx cm

5 cm

12 cm

36 cm

b

M

P

RQ

63 cm

16 cm

87 cm

y cm

cG

K

L

M

16 cm

30 cm

20 cm

x cm

Example 15

Estimating square rootsThe square numbers can be used to estimate square roots.

1 Examine these examples:a Estimate .

7 is between the square numbers 4 and 9 (22 and 32 respectively).

Now, because 7 is closer to 32 than to 22, will be closer to 3 than to 2.

An estimate is ≈ 2.6. (Actual answer = 2.645 …)

b Estimate .

55 is between the square numbers 49 and 64 (72 and 82 respectively).

55 is closer to 72 than to 82, will be closer to 7 than to 8.

An estimate is ≈ 7.4. (Actual answer = 7.416 …)

2 Now estimate these square roots:

a b c d

e f g h

i j k l

Number 1 2 3 4 5 6 7 8 9 10 11 12

Square 1 4 9 16 25 36 49 64 81 100 121 144

7

7

7

55

55

55

8 18 28 111

80 31 12 65

75 29 40 126

Skillbank 6

SkillTest 6-01

Estimating square roots


Example 16

1 A wire supporting a tower is 20 m in length. It is attached to the ground 10 m from the base of the tower. Find how far the wire reaches up the tower, giving your answer to the nearest 0.1 m.

SolutionFirst use the given information to draw the diagram and the triangle. Let h represent how far the wire reaches up the tower.

202 = 102 + h2

400 = 100 + h2

400 − 100 = h2

300 = h2

h = = 17.3205≈ 17.3

∴ The wire reaches 17.3 m up the tower.

2 A screen door is reinforced by two metal braces crossing the door diagonally. Find (in metres) the length of metal used in the two braces on the door.

SolutionLet g represent the length of a brace.g2 = 1.952 + 1.052 g2 = 4.905g =

≈ 2.21 (correct to two decimal places)

∴ The length of metal required for two braces is 2 × 2.21 m = 4.42 m.

(Note: The two braces are the same length because the diagonals of a rectangle are equal.)

3 Ms Turco’s briefcase has a rectangular base of 47 cm × 32 cm. Can her 60 cm umbrella fit in the briefcase, assuming that it must lie flat on the base?

SolutionIf the diagonal of the base of the briefcase is greater than or equal to 60 cm, then Ms Turco’s umbrella will fit in it.

Call the length of the diagonal x.x2 = 472 + 322 x2 = 3233x =

= 56.8594 …

∴ x < 60

Since the diagonal of the base of the briefcase is less than the length of the umbrella, the umbrella will not fit into the briefcase.

The diagram The triangle

20 m wire20 m

10 m 10 m

h m

300

changed to metres1.95 m

195 cm

1.05 m

105 cm

g m

4.905

47 cm32 cm

32 cm

x cm47 cm

3233


1 The size of a television screen is described by the length of its diagonal. So a 48 cm set has a diagonal length of 48 cm. Find the size of each of these TV screens, giving your answer correct to the nearest centimetre.

2 Find the length (l) of a ladder which reaches 2.1 m up a wall when the foot of the ladder is 1.4 m from the bottom of the wall. Give your answer to the nearest centimetre.

3 Sam takes a shortcut from his home to the shop. Find:a the length (to the nearest metre) of the shortcutb the distance Sam saves by not going the long way to

the shop.

4 The road from Arrowville to Bigtown is cut by floodwater, so Alison must travel from Arrowville to Bigtown via Cranjie. Find:a the distance (correct to one

decimal place) from Arrowville direct to Bigtown

b the extra distance Alison travels by taking the detour via Cranjie.

5 A flagpole is supported by a piece of wire which reaches 4 m up the flagpole. The wire is attached to the ground 2.4 m from the base of the flagpole. Find the length of the wire, correct to one decimal place.

6 On a cross-country ski course, checkpoints C and A are 540 m apart and checkpoints C and B are 324 m apart. Find:a the distance between checkpoints A and Bb the distance covered in one lap of the course,

giving your answer in kilometres, correct to one decimal place.

a b c

13 cm

11 cm

44 cm

31 cm

72 cm

128 cm

1.4 m

2.1 m

l m

Sam’s home

Shop127 m

84 m

shortcut

Bigtown

Cranjie

Arrowville4.2 km

3.8 kmflood

2.4 m

4 m

A

B

C

540 m

324 m


SkillBuilder 21-11Using



Intervals on the number planeSome terms and symbols

Points can be represented by an ordered pair of coordinates (x, y). For example, the point P has coordinates (2, 1).

7 The end of a 3.6 m plank touches the ground 3.5 metres from the bottom of a step as shown. How high (to the nearest centimetre) is the step?

8 For safety, a 2 m ladder should be placed no less than 27 cm, and no more than 90 cm, from a wall. Calculate, to the nearest centimetre:a the maximum height on the wall reached

by the ladderb the minimum height on the wall reached

by the ladder

9 A kite is attached to a 24 m piece of rope, as shown. The rope is held 1.2 m above the ground and covers a horizontal distance of 10 m. Find:a x, correct to one decimal placeb the height of the kite above the ground, correct to the

nearest metre.

10 A playground slide is made up of two right-angled triangles. Giving your answer to the nearest millimetre, find:a the height (h) of the slideb the length (l) of the slide.

Step

3.5 m

3.6 m

27 cm

2 m 2 m

90 cmMinimum distance

from wallMaximum distan

from wall

1.2 m10 m

24 m x m

l

h

4.5 m

1 m 5 m

ladder

slide

−4 4−3 3−2 2−1 1

−4

4

−3

3

−2

2

−1

1

y-axis

x-axis

QUADRANT2

QUADRANT1

QUADRANT4

QUADRANT3

P (2,1)

origin (0, 0)

The diagram below shows a Cartesian plane (or number plane).

The Cartesian number plane was named after René

Descartes (1596–1650).


Using intervals to form right-angled trianglesTwo points can be graphed to form the endpoints of an interval. If the interval is neither horizontal nor vertical, a right-angled triangle can be formed with that interval as the hypotenuse.

To do this:• draw a vertical side from the higher point, and• draw a horizontal side from the lower point.

Example 17

Plot these points on a number plane: A(5, 3) B(−4, −3) C(−2, 5) D(5, −5) a Draw the interval AD.b What is the length of the interval AD?

SolutionTo plot A(5, 3) go across 5 to the right from the origin and then up 3.

To plot B(−4, −3), go across 4 to the left from the origin and then down 3.

For C we go left 2 from the origin and up 5. For D we go right 5 from the origin and down 5.

a Draw a line from A to D. This is the interval AD.

b The length of AD is 8 units.

y

x

A

D

B

C

−4 4 5−3−5 3−2 2−1−1−2−3−4

−5

10

1

2

3

4

5

y

x

A

B

C

−4 4 5−3−5 3−2 2−1−1−2−3−4

−5

1

8 units

0

1

2

3

4

5

D

Example 18

Plot the points A(−3, 2) and B(5, −4) on a number plane.a With the interval AB as the hypotenuse, form a right-angled triangle.b What are the lengths of the horizontal and vertical sides of the triangle?


Solutiona A vertical side is drawn down from A

(the higher point) and a horizontal side is drawn to the left from B (the lower point). The two sides meet at C.

b The horizontal side BC is 8 units long. The vertical side AC is 6 units long.

y

x

A

BC

−4 4 5−3−5 3−2 2−1−1−2−3−4

−5

10

1

2

3

4

5

−6

1 For each of the following pairs of points:i plot the pair of points on a number plane ii draw the interval joining the points.a A(−3, 2) and B(5, −2) b C(1, 3) and D(−2, −1)c E(0, 5) and F(3, −2) d G(−3, 1) and H(2, 4)e J(−2, −4) and K(3, 0) f L(4, 3) and M(−3, −1)

2 Find the lengths of all the intervals shown on each of these number planes.

3 Plot the points T(2, −4) and W(−1, 2).a With the interval TW as the hypotenuse, form a right-angled triangle.b What are the lengths of the horizontal and vertical sides of the triangle?

y

x

A

−4 4 5−3−5 3−2 2−1−1−2−3−4

−5

10

1

2

3

4

5

−6 6

6

−6

D

P Q

B

M

N

y

x

F

−4 4 5−3−5 3−2 2−1−1−2−3−4

−5

10

1

2

3

4

5

−6 6

6

−6

D

P Y

K

H

E

W

T

ba


Example 18

WorksheetAppendix 4

Number plane grid paper


Finding the length of an intervalPythagoras’ theorem may be used to find the length of an interval on the Cartesian plane.

4 For each of the following pairs of points:i plot the pair of points on a number plane

ii draw the interval joining the two pointsiii with the interval as the hypotenuse, form a right-angled triangleiv find the lengths of the horizontal and vertical sides of the triangle.a B(−3, 0) and F(3, 4) b K(6, 3) and L(−4, −1) c P(2, −3) and N(−4, 5)d H(−3, −5) and O(0, 0) e W(0, 4) and E(−3, −2) f G(3, 1) and Q(−4, −4)g J(−3, −6) and R(3, 4) h A(1, −2) and D(−1, 5) i M(−5, 0) and R(0, −4)

Worksheet 6-07

Length of an interval

Example 19

The points A(−2, 1) and B(3, −2) are shown on the number plane on the right. Find, correct to one decimal place, the length of the interval AB.

SolutionForm a right-angled triangle ABC by drawing a vertical side from A (the higher point), and a horizontal side from B (the lower point) as shown.

∴ The point C is (−2, −2). So AC = 3 units, and BC = 5 units.

Using Pythagoras’ theorem:AB2 = 32 + 52

= 9 + 25AB2 = 34

∴ AB = ≈ 5.8 (correct to one decimal

place).

Note: The point C could also be (3, 1). Can you explain why?

−2

y

x

A

B

−2 2 3−1 1

−1

0

1

−3

−2

y

x

A

BC

−2 2 3−1 1

−1

0

1

−3

34

Spreadsheet 6-02

Interval length


Exact lengthThe answer of 5.8 units in Example 19 represents the approximate length of AB, correct to one decimal place. When the length was expressed using the irrational number , it was the exact length of AB.

34

Example 20

Find the length of the interval joining P(−4, 0) to Q(2, −1), expressing the answer in exact form.

SolutionFirst plot P and Q on a number plane and find another point (call it T) such that ∆PQT is right-angled at T. [Note that T could also be (2, 0) to make the right-angled triangle PTQ. It makes no difference to the length PQ].So PT = 1 unit, and TQ = 6 units.

PQ2 = 62 + 12

= 36 + 1PQ2 = 37PQ = (exact length)

y

T

x−4 −3 −2 2−1

−1

10

1

P(−4, 0)

Q(2, -1)37

1 For each interval shown on the number plane on the right, use Pythagoras’ theorem to find:i the exact length of the

intervalii its length correct to one

decimal place.

2 Find the distance between each pair of points A and B by plotting them on a Cartesian plane. Give your answer correct to one decimal place where necessary.a A(3, 2), B(6, −2) b A(−1, −1), B(4, 3) c A(−2, 5), B(1, 2)d A(3, 0), B(4, 6) e A(−5, 2), B(0, −1) f A(6, −1), B(−3, 2)g A(−2, −2), B(3, 2) h A(5, −4), B(0, 1) i A(3, 5), B(2, 4)j A(−1, 4), B(5, −4) k A(0, −5), B(5, 0) l A(4, −3), B(−4, 3)

y

x

c

−3 3−2 2−1

−1

−2

−3

10

1

2

4 5

3a

de

b


WorksheetAppendix 4

Number plane grid paper


3 a Find the exact length of PQ in the diagram on the right.

b Find the exact length of QR.c Find the exact length of PR.d What type of triangle is ∆PQR?e Find the perimeter of ∆PQR to one decimal

place.

y

x32−1

−1

−2

10

1

2

4

3

P

Q

R

Example 20

1 Name five hypotenuses in this diagram:

2 A square has an area of 144 cm2. Find the length of a diagonal, correct to one decimal place.

3 A pipe is placed inside a rectangular crate, as shown. The pipe is 1.6 m long. How much of the pipe extends over the top of the crate? Give your answer in metres, correct to two decimal places.

4 What is the maximum length of a steel rod that will fit inside a rectangular box whose base dimensions are 70 cm × 50 cm and whose height is 190 cm? How could the maximum length of the rod be determined?

5 The rectangle ABDE has an area of 160 m2. Find x if ∆BDC is an isosceles triangle.

A B

CD

E

F

1.4 m

45 cm

20 m

x m

A B

CD

E

Power plus


6 A square picture frame has a diagonal of 68 cm. Find the length of one side of the frame, giving your answer to the nearest centimetre.

7 Ribbon is to be sewn on a piece of material measuring 60 cm × 180 cm. The ribbon is sewn around the perimeter of ABCD, and then from A to E to F to C. Find (to the nearest 0.01 m) the total length of ribbon required.)

8 In the diagram on the right, find P, correct to one decimal place.

9 Here is one method of generating Pythagorean triads:• Choose any two numbers, a and b (where b > a).• Find the difference of their squares, b2 − a2.• Find twice the product of the two numbers, 2ab.• Find the sum of their squares, b2 + a2.Your results [(b2 − a2), (2ab), (b2 + a2)] form a Pythagorean triad.Generate a Pythagorean triad for each of the following pairs of numbers.a a = 15, b = 20 b a = 8, b = 26 c a = 4.5, b = 5.5

10 Using the fact that it is possible to find the end point of an interval starting at A(2, 3), such that the interval is units long.What are the coordinates of B?

B must be (2 + 1, 3 + 1) = (3, 4) and C must be (2 − 1, 3 + 1) = (1, 4).

a Find two more points (D and E) which are units from A(2, 3).

b Using then, starting at P(1, 4), the

point Q(5, 9) is units from P.

Find seven more points which are units from P(1, 4).

(Hint: )

68 cm

180 cm

60 cmA

B C

D

E

F G

20 cm

15 cm

12 cmp cm

2 12

12+=2 B(?, ?) C(?, ?)

A(2, 3)A(2, 3)

1 1

1 1

5

4

P(1, 4)

Q(5, 9)

41

2

41 16 25+ 42 52+= =

41

41

41 42 52+ 52 42+= =


Topic overview• Write, in your own words, what you have learned in this chapter.• Write which parts of this chapter were new to you.• Copy and complete:

The things I understand about Pythagoras’ theorem that I did not understand before are …The things I am still not confident in doing in this chapter are … (Give an example of each difficulty.)

• Copy and complete whichever applies to you:The steps I will take to overcome my problems with this chapter are …The sections of work that I found difficult in this chapter were …The sections of work that I found easy in this chapter were …

• Copy and complete the overview summary below.

Language of mathsCartesian plane diagonal distance exact length

hypotenuse interval irrational number number plane

proof prove Pythagoras’ theorem Pythagorean triad

quadrilateral right-angled side square

square root test theorem unknown

1 What is the meaning of the word ‘hypotenuse’?

2 Which word is another name for ‘rule’?

3 A diagonal is an interval from one vertex of a shape to another non-adjacent vertex. What does ‘non-adjacent’ vertex mean?

4 What name is given to any set of three numbers that follows Pythagoras’ theorem?

5 Describe Pythagoras’ theorem in your own words.

Worksheet 6-08


crossword

Solve problems usingPythagoras’

theorem

Test whether triangleis right-angled

or not

Identifythehypotenuse asthe longest side

Writeout thecurrent form ofPythagoras’ theorem

Find the length of thehypotenuse

Find the length ofone of theshortersides

THEOREMRight-angled triangles

Find the distance between two points on the number plane

PY

TH

AG

OR

AS’

PYTHAGORAS’ THEOREM 251 CHAPTER 6

1 Name the hypotenuse in each of these triangles.

2 Calculate the length of the hypotenuse for each of these triangles. Give your answer correct to one decimal place where necessary.

3 Calculate the length of the unknown side of each of these triangles. Give your answer correct to one decimal place where necessary.

4 Decide whether each of these triangles is right-angled or not.

5 Find x, correct to two decimal places, in each of these quadrilaterals.

6 Can an umbrella 84 cm long fit flat inside a briefcase whose rectangular base measures 48 cm × 64 cm?

7 Julie walks 6 km due east from a starting point P, while Jane walks 4 km due south from P. How far are Julie and Jane apart? (Give your answer correct to one decimal place.)

8 Plot the points P(−3, −2) and W(1, 5) on a number plane.a Draw the interval PW.b Use the interval PW as the hypotenuse, and form a right-angled triangle.c Write the lengths of the horizontal and vertical sides of the triangle.d Use Pythagoras’ theorem to find the length of PW (correct to one decimal place).

Chapter 6 Review

Ex 6-02

A

B

C

P

n

q

E

F

G

a b c

Ex 6-04

33 cm

44 cm

15 mm

112 mm

140 mm

105 mm

7.6 m

6.4

a b c d

Ex 6-05

56 mm

70 mm

30 mm

34 mm 150 mm

65 cm

63 cm87 mm

a b c d

Ex 6-06

45 51

20

35

37

12

69 115

92

a b c

Ex 6-0719.4 cm

11.4 cm

27.6 cm

x cm

x cm

16 cm

7 cm

24 cmba

Ex 6-08

Ex 6-08

W

N

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Ex 6-09

Topic testChapter 6

Measurement Pythagoras’ theorem - Spark Pythagoras’ theorem 6 ... 5aIf p2 = 36, ﬁnd p if p is...

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Transcript of Measurement Pythagoras’ theorem - Spark Pythagoras’ theorem 6 ... 5aIf p2 = 36, ﬁnd p if p is...